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For $\theta \in \left]0,\pi\right[$ and $n \in \mathbb{N}^{*}$, I'm asked whether the convergence of $$ a_n=\left(n! \prod_{k=1}^{n}\sin\left(\frac{\theta}{k}\right)\right)^{1/n} $$ What I've tried is to transform (if i'm right) $a_n$ into $$ \ln\left(a_n\right)=\frac{1}{n}\sum_{k=1}^{n}\ln\left(k\sin\left(\frac{\theta}{k}\right)\right) $$ I've proven that $$ \ln\left(k\sin\left(\frac{\theta}{k}\right)\right)=\ln\left(\theta\right)-\frac{\theta^2}{6k^2}+o\left(\frac{1}{k^2}\right) $$ So I found convergence in the case $\theta=1$, but with divergence of the series when $\theta \ne 1$ I can't conclude about the convergence of $\ln\left(a_n\right)$.

Furthermore, is it possible to find the limit of $a_n$ ?

Atmos
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  • Still when $k \rightarrow +\infty$ we can deduce whether $\sum \ln\left(\dots\right)$ converges or not, if it does, then we have convergence of $a_n$ – Atmos Sep 12 '19 at 20:35

1 Answers1

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Since $\lim\limits_{k\rightarrow +\infty}{\ln\left(k\sin\left(\frac{\theta}{k}\right)\right)=\ln\theta}$, by Cesaro's theorem you have that $$ \lim\limits_{n\rightarrow +\infty}\frac{1}{n}\sum_{k=1}^n{\ln\left(k\sin\left(\frac{\theta}{k}\right)\right)}=\ln\theta $$

Tuvasbien
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